# Given the sequence  { 6,18,54,162,... } , find the next three terms and an expression for the n^(th) term?

Feb 21, 2017

The relationship is that the ${n}^{t h}$ term is given by (assuming we start the sequence at $n = 1$:

${u}_{n} = 6 \times {3}^{n - 1}$

the next three terms are:

$486 , 1458 , 4374$

#### Explanation:

The sequence:

$\left\{6 , 18 , 54 , 162 , \ldots\right\}$

clearly is not linear or quadratic as the terms increase too rapidly, we note that they are all even and factors of $6$, so we let us see how factoring out $6$ helps to establish the pattern:

$\left\{6 , 6 \times 3 , 6 \times 9 , 6 \times 27 , \ldots\right\}$

and we can now see that the factor of $6$ is multiplied by a power of $3$, giving us:

$\left\{6 , 6 \times {3}^{1} , 6 \times {3}^{2} , 6 \times {3}^{3} , \ldots\right\}$

And we also now that ${3}^{0} = 1$, so we can modify the first term as follows:

$\left\{6 \times {3}^{0} , 6 \times {3}^{1} , 6 \times {3}^{2} , 6 \times {3}^{3} , \ldots\right\}$

And then we can see that the pattern is established and that the ${n}^{t h}$ term is given by (assuming we start the sequence at $n = 1$:

${u}_{n} = 6 \times {3}^{n - 1}$

So let us check that this works, and then form the next three terms:

$n = 1 \implies {u}_{1} = 6 \times {3}^{0} = 6$
$n = 2 \implies {u}_{2} = 6 \times {3}^{1} = 18$
$n = 3 \implies {u}_{3} = 6 \times {3}^{2} = 54$
$n = 4 \implies {u}_{4} = 6 \times {3}^{3} = 162$

$n = 5 \implies {u}_{5} = 6 \times {3}^{4} = 486$
$n = 6 \implies {u}_{6} = 6 \times {3}^{5} = 1458$
$n = 7 \implies {u}_{7} = 6 \times {3}^{6} = 4374$